A 1000-kg automobile moving with a speed of 24 m/s relative to the road collides with a 500-kg automobile initially at rest. If the two stick together, what is the velocity in m/s of the two cars after the collision according to an observer in a truck moving 10 m/s in the same direction as the moving cars?
A) 6
B) 11
C) 24
D) 26
E) 31

Question

Quizplus · Accepted Answer

Momentum of the system is conserved. The total momentum before the collision is:$$P_i = (1000 	ext{ kg})(24 	ext{ m/s}) + (500 	ext{ kg})(0 	ext{ m/s}) = 24000 	ext{ kg m/s}$$The total momentum after the collision is:$$P_f = (1500 	ext{ kg})v_f$$where $v_f$ is the velocity of the two cars after the collision.Since momentum is conserved, we have:$$P_i = P_f$$$$24000 	ext{ kg m/s} = (1500 	ext{ kg})v_f$$Solving for $v_f$, we get:$$v_f = \frac{24000 	ext{ kg m/s}}{1500 	ext{ kg}} = 16 	ext{ m/s}$$To an observer in a truck moving 10 m/s in the same direction as the moving cars, the velocity of the two cars after the collision is:$$v_{f,obs} = v_f + v_{truck} = 16 	ext{ m/s} + 10 	ext{ m/s} = 26 	ext{ m/s}$$Therefore, the correct answer is A.

A 1000-Kg Automobile Moving with a Speed of 24 M/s